Aus Online Mathematik Brückenkurs 1
(Unterschied zwischen Versionen)
Version vom 14:40, 22. Okt. 2008
Before we even start thinking about transforming \displaystyle \log_2 and \displaystyle \log_3 to ln, we use the log laws
|
| \displaystyle \begin{align}
\log a^b &= b\cdot\log a\,,\\[5pt]
\log (a\cdot b) &= \log a+\log b\,,
\end{align}
|
|
to simplify the expression
|
| \displaystyle \begin{align}
\log_{3}\log _{2}3^{118}
&= \log_{3}(118\cdot\log_{2}3)\\[5pt]
&= \log_{3}118 + \log_{3}\log_{2}3\,\textrm{.}
\end{align}
|
|
With help of the relation \displaystyle 2^{\log_{2}x} = x and \displaystyle 3^{\log_{3}x} = x and taking the natural logarithm , we can express \displaystyle \log_{2} and \displaystyle \log_{3} using ln,
|
| \displaystyle \log_{2}x=\frac{\ln x}{\ln 2}\quad and \displaystyle \quad\log_{3}x = \frac{\ln x}{\ln 3}\,\textrm{.}
|
|
The two terms \displaystyle \log_3 118 and \displaystyle \log_3\log_2 3 can therefore be written as
|
| \displaystyle \log_{3}118 = \frac{\ln 118}{\ln 3}\quad and \displaystyle \quad\log_{3}\log_{2}3 = \log_{3}\frac{\ln 3}{\ln 2}\,,
|
|
where we can simplify the last expression further with the logarithm law, log (a/b) = log a – log b, and then transform \displaystyle \log _{3} to ln,
|
| \displaystyle \begin{align}
\log_{3}\frac{\ln 3}{\ln 2}
&= \log_{3}\ln 3 - \log_{3}\ln 2\\[5pt]
&= \frac{\ln\ln 3}{\ln 3} - \frac{\ln\ln 2}{\ln 3}\,\textrm{.}
\end{align}
|
|
In all, we thus obtain
|
| \displaystyle \log_{3}\log_{2}3^{118} = \frac{\ln 118}{\ln 3} + \frac{\ln \ln 3}{\ln 3} - \frac{\ln\ln 2}{\ln 3}\,\textrm{.}
|
|
Input into the calculator gives
|
| \displaystyle \log_{3}\log_{2}3^{118}\approx 4\textrm{.}762\,\textrm{.}
|
|
Note: The button sequence on the calculator will be:
